2(3y-4)+8=7-2(3+2y)

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Solution for 2(3y-4)+8=7-2(3+2y) equation: 



2(3y-4)+8=7-2(3+2y)

We move all terms to the left:

2(3y-4)+8-(7-2(3+2y))=0

We add all the numbers together, and all the variables

2(3y-4)-(7-2(2y+3))+8=0

We multiply parentheses

6y-(7-2(2y+3))-8+8=0



We calculate terms in parentheses: -(7-2(2y+3)), so:

7-2(2y+3)

determiningTheFunctionDomain
-2(2y+3)+7

We multiply parentheses

-4y-6+7

We add all the numbers together, and all the variables

-4y+1

Back to the equation:

-(-4y+1)



We add all the numbers together, and all the variables

6y-(-4y+1)=0

We get rid of parentheses

6y+4y-1=0

We add all the numbers together, and all the variables

10y-1=0

We move all terms containing y to the left, all other terms to the right

10y=1

y=1/10

y=1/10

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